ON FIBONACCI NUMBERS WITH FEW PRIME DIVISORS
YANN BUGEAUD, FLORIAN LUCA, MAURICE MIGNOTTE, SAMIR SIKSEK
Abstract. If n is a positive integer, write Fn for the nth Fibonacci number,
and ω(n) for the number of distinct prime divisors of n. We give a description
of Fibonacci numbers satisfying ω(Fn ) ≤ 2. Moreover, we prove that the
inequality ω(Fn ) ≥ (log n)log 2+o(1) holds for almost all n. We conjecture that
ω(Fn ) log n for composite n, and give a heuristic argument in support of
this conjecture.
1. Introduction
Let Fn be the nth Fibonacci number and Ln be the nth Lucas number. In a
previous paper [3], the following result was proved.
Theorem 1. The only solutions to the equation
Fn = y m ,
m≥2
are given by n = 0, 1, 2, 6 and 12 which correspond respectively to Fn = 0, 1, 1, 8
and 144. Moreover, the only solutions to the equation
Ln = y p ,
m≥2
are given by n = 1 and 3 which correspond respectively to Ln = 1 and 4.
The proof involves an intricate combination of Baker’s method and the modular
method; it also needs about one week of computer verification using the systems
pari [1] and magma [2]. In this paper we use the above theorem, together with wellknown results of Carmichael and Cohn to give a description of Fibonacci numbers
with at most two distinct prime divisors.
If n is a positive integer, write ω(n) for the number of distinct prime divisors
of n. We prove that the inequality ω(Fn ) ≥ (log n)log 2+o(1) holds for almost all n.
We conjecture that ω(Fn ) log n for composite n, and give a heuristic argument
in support of this conjecture.
2. The Theorems of Carmichael and Cohn
We need the following two celebrated results on Fibonacci numbers. The first is
due to Carmichael [4] and the second to Cohn [5].
Theorem 2. Let n > 2 and n 6= 6, 12 then Fn has a prime divisor which does not
divide any Fm for 0 < m < n; such a prime is called a primitive divisor of Fn .
Date: February 13, 2005.
2000 Mathematics Subject Classification. Primary 11B39, Secondary 11K65.
Key words and phrases. Fibonacci numbers, arithmetic functions, prime divisors.
F. Luca’s work is funded in part by grants SEP-CONACyT 37259-E and 37260-E.
S. Siksek’s work is funded by a grant from Sultan Qaboos University (IG/SCI/DOMS/02/06).
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YANN BUGEAUD, FLORIAN LUCA, MAURICE MIGNOTTE, SAMIR SIKSEK
Theorem 3. Let 0 < m < n and suppose that the product Fm Fn is a square, then
(m, n) = (1, 2), (1, 12), (2, 12) or (3, 6).
3. Fibonacci numbers with ω(Fn ) ≤ 2
Write ω(m) for the number of distinct prime factors of m. Notice that ω(Fn ) > 0
for n > 2 and that for integers m and n, with 1 < m < n and mn 6= 6, 12, we have
(1)
ω(Fmn ) > ω(Fm ) + ω(Fn ),
if (m, n) = 1,
because first Fm and Fn both divide Fmn and are coprime when the indices are
coprime, and secondly Theorem 2 implies that there is a prime number dividing
Fmn which does not divide the product Fm Fn .
Lemma 3.1. Suppose n is a positive integer and and ω(Fn ) ≤ 2. Then either
n = 1, 2, 4, 8, 12 or n = `, 2`, `2 for some odd prime number `.
Proof. The Lemma follows straightforwardly from the fact that ω(Fn ) ≥ 3 in the
following cases:
(a) if 16 | n, or 24 | n, or
(b) if pq | n where p, q are distinct odd primes, or
(c) if `3 | n or 2`2 | n or 4` | n for some odd prime `, unless n = 12.
These may be verified using a combination of Theorem 2 and inequality (1). For
example, if 4` | n where ` 6= 3 is an odd prime then
ω(Fn ) ≥ ω(F4` ) > ω(F2` ) > ω(F` ) ≥ 1,
where the strict inequalities follow from Theorem 2. Thus ω(Fn ) ≥ 3.
If n = 2` with ` > 3 then Fn = F` L` and ω(Fn ) = 2 implies that F` and L` are
prime numbers (because of Theorem 1).
Consider now the case n = `2 with ω(Fn ) = 2. Then ω(F` ) = 1 and, by
Theorem 1, F` = p = prime. If ` = 5 then F` = 5 and F25 = 52 × 3001. If ` 6= 5
then p does not divide Fn /F` so that Fn = pq t , say. By Theorem 3, the product
F` Fn = p2 q t is not a square, thus t is odd.
We have proved the following result.
Theorem 4. The only solutions to the equation
ω(Fn ) = 1
are given by n = 4, n = 6 or n is an odd prime number for which Fn is prime.
The only solutions to the equation
ω(Fn ) = 2
are given for even n by n = 8 or n = 12 or n = 2` where ` is some odd prime
number for which Fn = F` L` where F` and L` are both prime numbers. For odd n,
the only possible cases are n = ` or n = `2 . Moreover if ω(Fn ) = 2 and n = `2 then
F` must be prime, say F` = p; and if ` 6= 5, then
Fn = pq t
where q is a prime number distinct from p, and the exponent t is odd.
A short computer search reveals examples of all of the above possibilities. Thus
FIBONACCI NUMBERS WITH FEW PRIME DIVISORS
3
• examples of ω(Fn ) = 1 with n prime are: F5 = 5, F7 = 13, F11 = 89,
F13 = 233, . . .
• examples of ω(Fn ) = 2 with n = 2` are: F10 = 55, F14 = 13 × 29, F22 =
89 × 199, F26 = 233 × 521, . . .
• examples of ω(Fn ) = 2 with n = `, `2 are: F9 = 34, F19 = 37 × 113, F25 ,
F31 = 557 × 2417, . . . , F121 = 89 × 97415813466381445596089, . . .
Remark. Lemma 3.1 can also be proved using some results from [7]. For example,
one of the results from [7] is that, if τ (m) denotes the number of divisors of the
positive integer m, then we have τ (Fn ) ≥ Fτ (n) for any positive integer n, with
equality only for n = 1, 2, 4. Assuming that ω(Fn ) ≤ 2, one gets immediately that
(up to a few exceptions) τ (n) ≤ 4, giving that n is the product of two distinct
prime numbers or the square of a prime number, which together with Theorem 2
yields the conclusion of Lemma 3.1.
4. Fibonacci numbers rarely have few prime factors
Theorem 4 in effect tells us that ω(Fn ) ≥ 3 for almost all n. We mean by this
that the set of integers for which this inequality holds has density 1. Indeed, much
more is true. For example, by the following theorem, ω(Fn ) ≥ C holds for almost
all n, whatever the value of C.
Theorem 5. The inequality ω(Fn ) ≥ (log n)log 2+o(1) holds for almost all n.
Proof. By Theorem 2, we know that if a divisor d of n is not 1, 2, 6 or 12 then Fd
has a primitive prime factor. This translates immediately in saying that
ω(Fn ) ≥ τ (n) − 4,
where τ (n) is the total number of divisors of n. Certainly, τ (n) ≥ 2ω(n) . Since
ω(n) = (1 + o(1)) log log n for almost all n, the desired inequality follows.
5. Heuristic results
Given the present state of knowledge in analytic number theory, it seems unrealistic to obtain precise results as to whether each of the possibilities from Theorem 4
occurs finitely many times or infinitely often. There is, however, a standard heuristic argument which suggests that there are infinitely many primes ` with F` prime,
but only finitely many primes ` with ω(F2` ) = 2 or ω(F`2 ) = 2.
The heuristic argument goes as follows (compare with [6, page 15]): from the
Prime Number Theorem, the ‘probability’ that a positive integer m is prime is
1/ log m. Thus the ‘expected number’ of primes ` such that F` is also prime is
X 1
X
1
≥A
log F`
`
` is prime
` is prime
for some positive constant A. Since this last series diverges (albeit very slowly), it
is reasonable to guess that there are infinitely prime F` .
Applying the same heuristic argument suggests that there are only finitely many
primes ` with ω(F2` ) = 2 or ω(F`2 ) = 2. For example, the ‘expected number’ of
primes ` with ω(F2` ) = 2 is
X
X 1
1
≤B
<∞
log F` × log L`
`2
` is prime
` is prime
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YANN BUGEAUD, FLORIAN LUCA, MAURICE MIGNOTTE, SAMIR SIKSEK
where B is some positive constant.
In fact we can go even further. We conjecture the following.
Conjecture 5.1. ω(Fn ) log n holds for all composite positive integers n.
In order to ‘justify’ this conjecture, let us make the following heuristic principle.
Heuristic 5.2. Let k : N → N be any function. Let A be a subset of positive
integers such that there is no algebraic reason for a ∈ A to have more than k(a)
prime factors. Assume that for every a ∈ A there exists a proper divisor of a, let
us call it ã > 1, such that the greatest common divisor of ã and a/ã has at most
one prime factor and the series
X
X
1
1
(log log ã)k1 −1 (log log(a/ã))k2 −1
a∈A k1 +k2 ≤k(a)+1
(k1 − 1)! (k2 − 1)!
log ã × log(a/ã)
is convergent. Then ω(a) ≤ k(a) should hold only for finitely many a ∈ A.
Heuristic 5.2, is based on the fact that the ‘probability’ for a positive integer n
to have k distinct prime factors is
k−1
1
(log log n)
(k − 1)!
log n
.
So, if a ∈ A and ω(a) = k ≤ k(a), then there exist nonnegative integers k1 and k2
such that ω(ã) = k1 and ω(a/ã) = k2 . Furthermore, since the greatest common
divisor of ã and a/ã has at most one prime factor, it follows that either k1 + k2 =
k ≤ k(a) or k1 + k2 = k + 1 ≤ k(a) + 1. Thus, the above sum represents just
the sum of the ‘probabilities’ that ω(ã) = k1 and ω(a/ã) = k2 assuming that such
events are independent.
5.1. Conjecture 5.1 follows from Heuristic 5.2. Let n be a composite integer.
Observe first that, by Theorem 2, if a divisor d of n is different from 1, 2, 6 or
12, then Fd has a prime factor not dividing Fd1 for any positive integer d1 < d.
Consequently, if n has at least 0.3 log n divisors, then we have
ω(Fn ) ≥ max{1, 0.3 log n − 6} log n.
(2)
Let A be the set of Fibonacci numbers Fn , where n runs through the composite
integers having less than 0.3 log n divisors. Let n be composite and write m for the
largest proper divisor of n. Clearly, m = n/p(n), where p(n) is the smallest prime
factor of n. Note that m ≥ n1/2 . It is known that gcd(Fm , Fn /Fm )|p(n), therefore
the two numbers Fm and Fn /Fm share at most one prime factor. Fix k and let k1
and k2 be such that k1 + k2 = k. One checks immediately that both inequalities
Fm < em and Fn /Fm < en−m hold. Thus, we get
X
1
k −1
k −1
(log log Fm ) 1 (log log(Fn /Fm )) 2
(k1 − 1)!(k2 − 1)!
k1 +k2 =k
≤
1
(k − 2)!
X
k1 +k2 =k
k−2
k −1
(log m)k1 −1 (log(n − m)) 2
k1 − 1
1
k−2
=
(log m + log(n − m))
<
(k − 2)!
2e log n
k−2
k−2
,
FIBONACCI NUMBERS WITH FEW PRIME DIVISORS
5
where for the last inequality above we used Stirling’s formula. Using the inequalities
log Fm m ≥ n1/2 and log(Fn /Fm ) (n − m) n (because m divides n), it
follows that
k−2
k −1
k −1
X
(log log Fm ) 1 (log log(Fn /Fm )) 2
2e log n
1
.
3/2
(k1 − 1)!(k2 − 1)! log Fm × log(Fn /Fm )
k−2
n
k1 +k2 =k
For a fixed y, the function x 7−→ (2ey/x)x is increasing for x < 2y. Thus, taking
k(Fn ) = c0 log n, where c0 < 2 is some constant, we get that
k −1
k −1
X
(log log Fm ) 1 (log log(Fn /Fm )) 2
k1 +k2 ≤k(Fn )
X
k≤k(Fn )
(k1 − 1)!(k2 − 1)! log Fm × log(Fn /Fm )
1
n3/2
2e log n
k−2
k−2
log n
3/2
n
2e
c0
c0 log n
.
Choosing c0 such that c0 log(2e/c0 ) = c1 < 1/2 (we can choose, say, c0 = 0.3), we
get that the right hand side of the above inequality is
log n
3/2−c
1
n
and summing up over n we get a convergent series. Hence, Heuristic 5.2 with
a = Fn and ã = Fm for Fn in A implies that ω(Fn ) < 0.3 log n holds only for
finitely many composite integers n, which, by (2) and the definition of A, implies
that ω(Fn ) log n holds for all composite integers n. Actually, our choice of A
takes into account an algebraic reason for which Fn has more than 0.3 log n divisors.
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Yann Bugeaud, Université Louis Pasteur, U. F. R. de mathématiques, 7, rue René
Descartes, 67084 Strasbourg Cedex, France
E-mail address: [email protected]
Florian Luca, Instituto de Matemáticas, UNAM, Ap. Postal 61-3 (Xangari), C.P.
58180, Morelia, Michoacán, Mexico
E-mail address: [email protected]
Maurice Mignotte, Université Louis Pasteur, U. F. R. de mathématiques, 7, rue René
Descartes, 67084 Strasbourg Cedex, France
E-mail address: [email protected]
Samir Siksek, Department of Mathematics and Statistics, College of Science, Sultan
Qaboos University, P.O. Box 36, Al-Khod 123, Oman
E-mail address: [email protected]